Two taps can separately fill a cistern in 20 minutes and 25 minutes. Both taps are open for 10 minutes after which the slower one is closed. How long will it take to fill the remaining portion by the other tap alone?

$$ work done = efficiency \times time $$

Two taps P1 and P2  can separately fill a cistern in 20 minutes and 25 minutes.

work done = LCM( 20,25 ) =100

efficiency of P1 = $$ \frac{100}{20} = 5 $$

efficiency of P2 = $$ \frac{100}{25} = 4 $$

Both taps are open for 10 minutes after which the slower one is closed.

$$ work done = ( efficiency of P1 + efficiency of P2 ) \times 10 + efficiency of P1 \times x $$

where x = time required by P1 to fill the remaining portion by the other tap alone

substituting

$$ 100 = ( 5 + 4) \times 10 + 5x $$

solving x = 2 minutes